If BC is a diameter of a circle of circle of centre O and OD is perpendicular to the chord $¯ ¯¯¯¯¯¯ ¯ A B$ of a circle then $C A \equiv$ ________

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Solution

$∵ O B = O C = r a d i u s = r$
Join $O A$ which is radius i.e $O A = r$
$O D ⊥ A B$ in isosceles triangle $A O$ and so divide $¯ ¯¯¯¯¯¯ ¯ A B$ into two square sides $¯ ¯¯¯¯¯¯¯ ¯ A D$ and $\overline{BD}$
By applying mid point theorem in $△ A B C$ ,
$\Rightarrow O D ∥ A C$
$\Rightarrow ¯ ¯¯¯¯¯¯¯ ¯ O D = \frac{1}{2} ¯ ¯¯¯¯¯¯ ¯ A C$
$\Rightarrow ¯ ¯¯¯¯¯¯ ¯ A C = 2 ¯ ¯¯¯¯¯¯¯ ¯ O D$

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If BC is a diameter of a circle of circle of centre O and OD is perpendicular to the chord ¯¯¯¯¯¯¯¯AB of a circle then CA≡________

If BC is a diameter of a circle of circle of centre O and OD is perpendicular to the chord $¯ ¯¯¯¯¯¯ ¯ A B$ of a circle then $C A \equiv$ ________